Certification Problem
Input (TPDB TRS_Standard/Zantema_05/jw50)
The rewrite relation of the following TRS is considered.
|
f(f(a,f(a,a)),x) |
→ |
f(x,f(f(a,a),a)) |
(1) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Uncurrying
We uncurry the binary symbol
f
in combination with the following symbol map which also determines the applicative arities of these symbols.
| a |
is mapped to |
a, |
a1(x1), |
a2(x1, x2) |
There are no uncurry rules.
No rules have to be added for the eta-expansion.
Uncurrying the rules and adding the uncurrying rules yields the new set of rules
|
a2(a1(a),x) |
→ |
f(x,a2(a,a)) |
(4) |
|
f(a,y1) |
→ |
a1(y1) |
(2) |
|
f(a1(x0),y1) |
→ |
a2(x0,y1) |
(3) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a) |
= |
1 |
|
weight(a) |
= |
1 |
|
|
|
| prec(a1) |
= |
0 |
|
weight(a1) |
= |
1 |
|
|
|
| prec(a2) |
= |
3 |
|
weight(a2) |
= |
0 |
|
|
|
| prec(f) |
= |
2 |
|
weight(f) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
a2(a1(a),x) |
→ |
f(x,a2(a,a)) |
(4) |
|
f(a,y1) |
→ |
a1(y1) |
(2) |
|
f(a1(x0),y1) |
→ |
a2(x0,y1) |
(3) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.